Find the adjoint of each of the following matrices: (i) $\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]$ (ii) $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices.

Find the adjoint of each of the following matrices:
(i) $\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]$
(ii) $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices.

Adjoint and Inverse of a Matrix | Exercise 7.1 | Maths | RD Sharma

Solution:

(i) Suppose

$A=\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]$

Cofactors of $A$ are

$C_{11}=4$

$C_{12}=-2$

$C_{21}=-5$

$C_{22}=-3$

Since, adj $A=\left[\begin{array}{ll}C_{11} & C_{12} \\ C_{21} & C_{22}\end{array}\right]^{1}$

$(\operatorname{adj} A)=\left[\begin{array}{cc}4 & -2 \\ -5 & -3\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{cc}4 & -5 \\ -2 & -3\end{array}\right]$

Now, $(\operatorname{adj} A) A=\left[\begin{array}{cc}4 & -5 \\ -2 & -3\end{array}\right]\left[\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right]=\left[\begin{array}{cc}-12-10 & 20-20 \\ 6-6 & -10-12\end{array}\right]$

$(\operatorname{adj} A) A=\left[\begin{array}{cc}-22 & 0 \\ 0 & -22\end{array}\right]$

And, $|\mathrm{A}| \mathrm{I}=\left|\begin{array}{cc}-3 & 5 \\ 2 & 4\end{array}\right|\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=(-22)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-22 & 0 \\ 0 & -22\end{array}\right]$

$\begin{aligned} &\text { Also, } A(\operatorname{adj} A)=\left[\begin{array}{cc} -3 & 5 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} 4 & -5 \\ -2 & -3 \end{array}\right]=\left[\begin{array}{cc} -12-10 & 20-20 \\ 6-6 & -10-12 \end{array}\right] \\ &A(\operatorname{adj} A)=\left[\begin{array}{cc} -22 & 0 \\ 0 & -22 \end{array}\right] \\ &\text {As a result, }(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A) \end{aligned}$

(ii) Suppose

$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

Therefore cofactors of $A$ are

$\mathrm{C}_{11}=\mathrm{d}$

$C_{12}=-c$

$C_{21}=-b$

$C_{22}=a$

It is known that, $\operatorname{adj} A=\left[\begin{array}{ll}\mathrm{C}_{11} & \mathrm{C}_{12} \\ \mathrm{C}_{21} & \mathrm{C}_{22}\end{array}\right]^{\mathrm{T}}$

So by substituting these values we obtain,

$(\operatorname{adj} A)=\left[\begin{array}{cc}d & -c \\ -b & a\end{array}\right]^{T}$

$=\left[\begin{array}{cc}\mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a}\end{array}\right]$

Now, $\left(\right.$ adj A) $A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}a d-b c & b d-b d \\ -a c+a c & -b c+a d\end{array}\right]$

$(\operatorname{adj} A) A=\left[\begin{array}{cc}a d-b c & 0 \\ 0 & a d-b c\end{array}\right]$

And, $|\mathrm{A}| . \mathrm{I}=\left|\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{C} & \mathrm{d}\end{array}\right|\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=(\mathrm{ad}-\mathrm{b} \mathrm{c})\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\mathrm{ad}-\mathrm{bc} & 0 \\ 0 & \mathrm{ad}-\mathrm{bc}\end{array}\right]$

Also,

$\begin{aligned} &A(\operatorname{adj} A)=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]=\left[\begin{array}{cc} a d-b c & 0 \\ 0 & a d-b c \end{array}\right] \\ &\text {As a result, }(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A) \end{aligned}$

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